Question

A photon and an electron possesses same de-Broglie wavelength. Given that C$=$ speed of light and v$=$ speed of electron, which of the following relation is correct? (Here, $E_e=K.E$ of electron, $E_{Ph}=K.E$ of photon, $P_e=$ momentum of electron, $P_{ph}=$ momentum of photon).

• A. $\frac{P_e}{P_{Ph}}=\frac{C}{2v}$
• B. $\frac{E_e}{E_{Ph}}=\frac{C}{2v}$
• C. $\frac{E_{ph}}{E_e}=\frac{2c}{v}$
• D. $\frac{P_e}{P_{Ph}}=\frac{2c}{v}$

Answer 1

The correct option is C $\frac{E_{ph}}{E_e}=\frac{2c}{v}$

We have, ${\lambda }_{Ph}=\frac{h}{P_{Ph}}$ and ${\lambda }_e=\frac{h}{P_e}$
Given, ${\lambda }_{Ph}={\lambda }_e$
$\therefore$ We get, $P_{Ph}=P_e$
$\frac{h}{{\lambda }_{Ph}}=mv$
$\therefore \frac{hc}{{\lambda }_{Ph}}=mcv=\frac{1}{2}m{v}^2\left(\frac{2c}{v}\right)$
or $\frac{E_{Ph}}{E_e}=\frac{2c}{v}$.